How do I solve this?

Question

A cinema has two different prices for tickets: one price for children and one price for adults.

: 35 cents children's tickets plus 20 adults cost a total of 287.5

:25 cents children's tickets plus 30 adults tickets cost 307.50

How much does a single adult ticket cost, in dollars

Answer 1

Hi Eva N.,

The [cost of a child's ticket] times the [number of child tickets purchased] plus [the cost of an adult ticket] times the [number of adult tickets purchased] equals the total cost of all tickets purchased.

So the equation is:

[cost of a child's ticket]*[number of child tickets purchased] + [the cost of an adult ticket]*[number of adult tickets purchased] = total cost of all tickets purchased.

Let's let:

a = cost of a child's ticket

x = number of child tickets purchased

b = the cost of an adult ticket (((this is the answer we are looking for)))

y = number of adult tickets purchased

The equation becomes: a*x + b*y = total cost of all tickets purchase.

We can now write two equations with the information provided.

For equation(1) we know; a = .35, y = 20, and the total cost of all tickets purchase 287.50;

.35*x + b*20 = 287.50

For equation(2) we know; a = .25, y = 30, and the total cost of all tickets purchase 307.50;

.25*x + b*30 = 307.50

Since we are looking for (((b = the cost of an adult ticket))), we can use elimination. We can multiply 5 times equation(1), and 7 times equation(2), and subtract equation(1) from equation(2).

5*(.35*x + b*20 = 287.50)---->>> 1.75*x + b*100 = 1437.50

7*(.25*x + b*30 = 307.50)---->>> 1.75*x + b*210 = 2152.50

1.75*x + b*210 = 2152.50

- [1.75*x + b*100 = 1437.50]

---------------------------------------

b*110 = 715.00

b = 6.50

A single adult ticket cost of $6.50.

I hope this helps, Joe.

P.S.

The problem states; 'A cinema has two different prices for tickets: one price for children and one price for adults.', which is incorrect.

The problem should state; 'A cinema has three different prices for tickets: two prices for children and one price for adults.'

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Hi Eva N.,

You can multiply either equation by a negative multiplier, then you must add the two equations. I prefer to multiply by positive numbers, then subtract the smaller (b*) from the larger (b*) to keep my answer positive. Negative signs can be an easy way to make a mistake.

For example:

(-5)*(.35*x + b*20 = 287.50)---->>> -1.75*x - b*100 = -1437.50

7*(.25*x + b*30 = 307.50)---->>> 1.75*x + b*210 = 2152.50

Then add:

-1.75*x - b*100 = -1437.50

+[1.75*x + b*210 = 2152.50]

------------------------------------------

b*110 = 715.00

b = 6.50

Or:

5*(.35*x + b*20 = 287.50)---->>> 1.75*x + b*100 = 1437.50

(-7)*(.25*x + b*30 = 307.50)---->>> -1.75*x - b*210 = -2152.50

Then add:

1.75*x + b*100 = 1437.50

+[-1.75*x - b*210 = -2152.50]

--------------------------------------------

-b*110 = -715.00

b = 6.50

There are different ways to solve a problem. You need to find the way that suits you best.

If you still have questions please don't hesitate to ask me. I would like you to have a clear understanding.

Thanks, Joe.